| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Topic | Complex numbers 2 |
| Type | Solve polynomial equations with complex roots |
| Difficulty | Challenging +1.8 This is a challenging complex numbers problem requiring multiple sophisticated techniques: using |w|=1 to derive a constraint, manipulating the equation algebraically to find k (involving nested radicals), then analyzing the remaining roots using properties of polynomial equations and modulus inequalities. Part (ii) requires proof rather than calculation, demanding insight about how roots relate when one has |w|=1. This goes well beyond routine complex number exercises and requires extended reasoning across multiple concepts. |
| Spec | 4.02a Complex numbers: real/imaginary parts, modulus, argument4.02b Express complex numbers: cartesian and modulus-argument forms4.02i Quadratic equations: with complex roots |
| Answer | Marks | Guidance |
|---|---|---|
| (i) If \(w\) is a root with \( | z | =1\), then \(w = \cos\theta + \mathrm{i}\sin\theta\) Modelling M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \( | c | \leq 1 \Rightarrow \cos\theta = \frac{1}{2}(1-\sqrt{3})\) A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Then \(w^2 = a^2 - b^2 + \mathrm{i}\cdot 2ab\) and \(\frac{1}{w} = \frac{a-\mathrm{i}b}{a^2+b^2} = a - \mathrm{i}b\) (since \( | w | =1\)) B1 B1 |
| Answer | Marks | Guidance |
|---|---|---|
| However, \( | a | < 1 \Rightarrow a = \frac{1}{2}(1-\sqrt{3})\) or similar argument from \(a^2 = 1 - \frac{\sqrt{3}}{2}\) A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Then \( | \alpha | = 1 \Rightarrow |
(i) If $w$ is a root with $|z|=1$, then $w = \cos\theta + \mathrm{i}\sin\theta$ Modelling **M1**
Then $\cos 2\theta + \mathrm{i}\sin 2\theta - 2(\cos\theta - \mathrm{i}\sin\theta) + k\mathrm{i} = 0$ Use of de Moivre's theorem **M1**
$\cos 2\theta - 2\cos\theta = 0$ Considering real parts **M1**
$2c^2 - 2c - 1 = 0$ Solving **M1**
$|c| \leq 1 \Rightarrow \cos\theta = \frac{1}{2}(1-\sqrt{3})$ **A1**
$\sin\theta = \pm\sqrt{1 - \frac{1}{4}(4-2\sqrt{3})} = \pm\sqrt{\frac{1}{2}\sqrt{3}}$ **A1**
$k = -\sin 2\theta - 2\sin\theta\cos\theta$ or $-2\sin\theta(1+\cos\theta)$ Considering imaginary parts **M1**
$= \mp 2 \times \sqrt{\frac{1}{2}\sqrt{3}} \times \frac{1}{2}(3-\sqrt{3}) = (3-\sqrt{3})\sqrt{\frac{1}{2}\sqrt{3}}$ **Answer Given** **A1** [8]
**ALTERNATIVE I**
Let $w = a + \mathrm{i}b$ where $a^2 + b^2 = 1$
Then $w^2 = a^2 - b^2 + \mathrm{i}\cdot 2ab$ and $\frac{1}{w} = \frac{a-\mathrm{i}b}{a^2+b^2} = a - \mathrm{i}b$ (since $|w|=1$) **B1 B1**
giving $(a^2 - 2a - b^2) + \mathrm{i}(2ab + 2b + k) = 0$
$\Rightarrow a^2 - 2a - b^2 = 0$ **and** $2ab + 2b + k = 0$ **M1**
Using $b^2 = 1 - a^2$ in Real part $= 0$ and solving a quadratic in $a$: $2a^2 - 2a - 1 = 0$ **M1**
$\Rightarrow a = \frac{1}{2}(1 \pm \sqrt{3})$
However, $|a| < 1 \Rightarrow a = \frac{1}{2}(1-\sqrt{3})$ or similar argument from $a^2 = 1 - \frac{\sqrt{3}}{2}$ **A1**
Thus $b^2 = \frac{\sqrt{3}}{2}$ and $b = \pm\sqrt{\frac{\sqrt{3}}{2}}$ **A1**
Substituting for $a$ and $b$ in Im part $= 0$ **M1**
$\Rightarrow k = -2b(a+1) = \mp 2\cdot\sqrt{\frac{1}{2}\sqrt{3}} \cdot \frac{1}{2}(3-\sqrt{3})$
$= (3-\sqrt{3})\sqrt{\frac{1}{2}\sqrt{3}}$ (since told $k > 0$) **Answer Given** **A1**
(ii) If $w^3 + kiw - 2 = 0$ has roots $\alpha, \beta, \gamma$, then $\alpha\beta\gamma = 2$ **B1**
Then $|\alpha| = 1 \Rightarrow |\beta\gamma| = 2$ and at least one of $\beta$, $\gamma$ has magnitude $> 1$ **B1** [2]13 The complex number $w$ has modulus 1. It is given that
$$w ^ { 2 } - \frac { 2 } { w } + k \mathrm { i } = 0$$
where $k$ is a positive real constant.\\
(i) Show that $k = ( 3 - \sqrt { 3 } ) \sqrt { \frac { 1 } { 2 } \sqrt { 3 } }$.\\
(ii) Prove that at least one of the remaining two roots of the equation $z ^ { 2 } - \frac { 2 } { z } + k i = 0$ has modulus greater than 1 .
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2014 Q13 [8]}}